In the paper *On the Castelnuovo-Mumford regularity of the cohomology ring of a group*, Symonds describes the following space.

Let $G = (\mathbb{Z}/2\mathbb{Z})^2 = \{1,a,b,ab\}$ be an elementary abelian $2$-group, and let $Z = \{ z_1, \dots, z_6\}$ be a discrete space with six elements. Let $G$ act on $Z$ such that each rank one subgroup of $G$ is a stabilizer of some element of $Z$. If I'm not mistaken, this means that the action is for example given as follows: $$a \cdot z_1 = b \cdot z_1 = z_2;\\ a \cdot z_3 = z_4; \; b \cdot z_3 = z_3; \; b \cdot z_4 = z_4;\\ b \cdot z_5 = z_6; \; a \cdot z_5 = z_5; \; a \cdot z_6 = z_6.$$

Let $SZ$ be the suspension of $Z$, with the induced $G$-action.

**Question.** How to compute the equivariant cohomology $H^*_G(SZ)$ as an $H^*_G$-module (with coefficients $\mathbb{F}_2$)?

I believe I could compute the equivariant cohomology $H^*_G(Z)$, using the Mayer–Vietoris sequence for equivariant cohomology (it split as a disjoint union of three $G$-spaces), the Künneth theorem, and the known results about the equivariant cohomology of spaces with a free (resp. trivial) action. But unfortunately there's no suspension isomorphism for equivariant cohomology, as far as I can tell...

Of course I know the definition of equivariant cohomology; the classifying space $BG$ is $(\mathbb{RP}^\infty)^2$, and $H^*_G = \mathbb{F}_2[x,y]$ is a polynomial algebra on two variables of degree $1$. The total space of the universal bundle is $(S^\infty)^2$ with $a$ (resp. $b$) acting by the antipodal action on the first (resp. second) factor, and $H^*_G(SZ) = H^*(EG \times_G SZ)$. But this isn't exactly helpful, or at least I don't see how to compute the equivariant cohomology just from that.